James W. Brennan

Factoring a polynomial is the opposite process of multiplying polynomials.
Recall that when we factor a number, we are looking for prime factors that
multiply together to give the number; for example

6 = 2 × 3 , or
12 = 2 × 2 × 3.

When we factor a polynomial, we are looking for simpler polynomials that can
be multiplied together to give us the polynomial that we started with. You
might want to review multiplying polynomials if you are not completely clear on
how that works.

When we factor a polynomial,
we are usually only interested in breaking it down into polynomials that have integer
coefficients and constants.

The simplest type of factoring is when there is a factor common to every
term. In that case, you can factor out that common factor. What you are doing
is using the distributive law in reverse—you are sort of un-distributing the
factor.

Recall that the distributive law says

a(b + c) = ab + ac.

Thinking about it in reverse means that if you see ab + ac,
you can write it as a(b + c).

where the coefficients a, b, and c, are real numbers
(for simplicity we will only use integers, but in real life they could be any
real number). We are interested here in factoring quadratic trinomials with
integer coefficients into factors that have integer coefficients. Not all such
quadratic polynomials can be factored over the real numbers, and even fewer
into integers (they all can be factored of we allow for imaginary numbers and
rational coefficients, but we don't). Therefore, when
we say a quadratic can be factored, we mean that we can write the factors
with only integer coefficients.

If a quadratic can be factored, it will be the product of two first-degree
binomials, except for very simple cases that just involve monomials. For
example x2 by itself is a quadratic expression where the
coefficient a is equal to 1, and b and c are zero.
Obviously, x2 factors into (x)(x), but this is
not a very interesting case.

A slightly more complicated case occurs when only the coefficient c
is zero. Then you get something that looks like

2x2 + 3x

This can be factored very simply by factoring out (‘undistributing’) the
common factor of x:

2x2 + 3x = x(2x + 3)

The most general case is when all three terms are present, as in

x2 + 5x + 6

We look at two cases of this type. The easiest to factor are the ones where
the coefficient of x2 (which we are calling ‘a’) is
equal to 1, as in the above example. If a is not 1 then things get a
little bit more complicated, so we will begin by looking at a = 1
examples.

Coefficient of x2 is 1

Since the trinomial comes from multiplying two first-degree binomials, let’s
review what happens when we multiply binomials using the FOIL method. Remember
that to do factoring we will have to think about this process in reverse (you
could say we want to ‘de-FOIL’ the trinomial).

Suppose we are given

(x + 2)(x + 3)

Using the FOIL method, we get

(x + 2)(x + 3) = x2 + 3x + 2x + 6

Then, collecting like terms gives

(x + 2)(x + 3) = x2 + 5x + 6

Now look at this and think about where the terms in the trinomial came from.
Obviously the x2 came from x times x. The
interesting part is what happens with the other parts, the ‘+ 2’ and the ‘+ 3’.
The last term in the trinomial, the 6 in this case, came from multiplying
the 2 and the 3. Where did the 5x in the middle come from? We got the 5x
by adding the 2x and the 3x when we collected like terms.
We can state this as a rule:

If the coefficient of x2
is one, then to factor the quadratic you need to find two numbers that:

1. Multiply to give the constant term (which we
call c)

2. Add to give the coefficient of x (which
we call b)

This rule works even if there are minus signs in the quadratic expression
(assuming that you remember how to add and multiply positive and negative numbers).

The trinomials on the right are called perfect squares because they
are the squares of a single binomial, rather than the product of two different
binomials. A quadratic trinomial can also have this form:

(x + 3)2 = (x + 3)(x + 3) = x2 + 6x + 9

Notice that just as before the coefficient of x is the sum
3 + 3, and the constant term is the product 3 ´ 3.
One can also say that

The coefficient of x
is twice the number 3

The constant term is
the number three squared

In general, if a quadratic trinomial is a perfect square, then

oThe coefficient of x
is twice the square root of the constant term

Or to put it another way,

oThe constant term is the
square of half the coefficient of x

In symbolic form we can express this as

It is helpful to be able to recognize perfect square trinomials. We will see
them again when we talk about solving quadratic equations.

Coefficient of x2 is not 1

A quadratic is more difficult to factor when the coefficient of the squared
term is not 1, because that coefficient is mixed in with the other products
from FOILing the two binomials. There are two methods for attacking these:
either you can use a systematic guess-and-check method, or a method called
factoring by grouping. We will first look at the guess-and-check method (which
we could call factoring by grouping).

If you need to factor a trinomial such as

2x2 + x- 3,

you have to think about what combinations could give the 2x2
as well as the other two terms. In this example the 2x2 must
come from (x)(2x), and the constant term might come from either (-1)(3) or (1)(-3).
The hard part is figuring out which combination will give the correct middle
term. This gets messy because all those coefficients will be mixed in with the
middle term when you FOIL the binomials. To see what is going on, let’s see
what happens when we FOIL the following binomials:

What happened? There are several significant things to notice:

The leading term in
the trinomial (the 2x2) is just the product of the
leading terms in the binomials.

The constant term in
the trinomial (the -3) is the
product of the constant terms in the binomials (so far this is the same as
in the case where the coefficient of x2 is 1)

The middle term in the
trinomial (the x) is the sum of the outer and inner products, which
involves all the constants and coefficients in the binomials, in a
messy way that is not always obvious by inspection.

Because 1 and 2 are relatively simple and 3 is complicated, it makes sense
to think of the possible candidates that would satisfy conditions 1 and 2, and
then test them in every possible combination by multiplying the resulting
binomials to see if you get the correct middle term. This seems tedious, and
indeed it can be if the numbers you are working with have a lot of factors, but
in practice you usually only have to try a few combinations before you see what
will work. As a demonstration, let’s see how we would attack the example by
this method.

Given

2x2 + x- 3

We make a list of the possible factors of 2x2: The only
choice is (2x)(x).

Then we make a list of the possible factors of the constant term -3: it is either (1)(-3) or (-1)(3). (Notice
that since we need a negative number, one factor must be negative and the other
positive, but it doesn’t matter which one so we have to try it both ways).

The possible factors of the trinomial are the binomials that we can make out
of these possible factors, taken in every possible order. From these
possibilities, we see that the candidate binomials are:

(2x + 1)(x – 3)
(x + 1)(2x – 3)
(2x + 3)(x – 1)
(x + 3)(2x – 1)

If we start multiplying these out, we will find that the third one works,
and then we are finished. All you really need to check is to see if the sum of
the outer and inner multiplications will give you the correct middle term,
since we already know that we will get the correct first and last terms.

In short, the method is:

1.List all the possible ways to get the coefficient of x2
(which we call a) by multiplying two numbers

2.List all the possible ways to get the constant term (which we
call c) by multiplying two numbers

3.Try all possible combinations of these to see which ones give
the correct middle term

·Don’t forget that the number itself times 1 is a
possibility

·If the number (a or c) is negative,
remember to try the plus and minus signs both ways